Integrally Closed is Local Property
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Theorem
Let $A$ be an integral domain.
For a prime ideal $\mathfrak p$ of $A$, let $A_{\mathfrak p}$ denote the localization at $S = A \divides \mathfrak p$.
Then the following are equivalent:
- $(1): \quad A$ is integrally closed
- $(2): \quad A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$.
- $(3): \quad A_{\mathfrak m}$ is integrally closed for all maximal ideals $\mathfrak m$.
Proof
$(1)$ implies $(2)$
Let $\map Q R$ denote the field of quotients of an integral domain $R$.
We have by Localization Preserves Integral Closure that:
- $\map Q {A_{\mathfrak p}} = \map Q A$
Hence $A_{\mathfrak p}$ is integrally closed for all prime ideals $\mathfrak p$.
The validity of the material on this page is questionable. In particular: What is the point of considering field of quotients? The claim seems exactly Localization Preserves Integral Closure You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by resolving the issues. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Questionable}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |
$\Box$
$(2)$ implies $(3)$
This is true because a Maximal Ideal of Commutative and Unitary Ring is Prime Ideal.
$\Box$
$(3)$ implies $(1)$
This theorem requires a proof. In particular: Need some more results on localization first You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by crafting such a proof. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{ProofWanted}} from the code.If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page. |